(B) KInetic energy at P is equal to potential energy at Q
The idea here is that the energy that the ball has in point
So, the ball starts at a point
#E_P = m * g * p ->#potential energy at p meteres above the ground
Now, what would the ball need in order to go from a point
SInce it's being thrown, an initial velocity directed straight upward must be applied to the ball. This means that at point
Let's assume that the ball is being thrown with a velocity
#E_P = m * g * p + 1/2 * mv^2#
What will happen as the ball moves upward?
Since it's being acted upon by gravity, which always points downward, its initial velocity will decrease as the height of the ball increases.
This means that its kinetic energy will decrease as well. At its highest point, the ball's velocity will be equal to zero.
SInce the height at which the ball is located increases, its potential energy will increase as well. Let's assume that point
#E_Q = m * g * q + 1/2 * m * 0^2 = m * g * q#
Since the ball started from
Since the energy of the ball must be conserved in points
#E_P = E_Q#
#m * g * p + 1/2 * m * v^2 = m * g * q#
If you rearrange this equation, you'll get
#1/2 * m * v^2 = m * g * q - m * g * p#
#1/2 * m * v^2 = m * g * (q-p)#
This tells you that the kinetic energy that the ball had in point
If you take point
#1/2 * m * v^2 = m * g * q ->#kinetic energy at p is equal to potential energy at q